Lesson 2-1 Tangent and Velocity Problems

5-Minute Check on Algebra Transparency 1-1 Click the mouse button or press the Space Bar to display the answers. 1.6x + 45 = 18 – 3x 2.x 2 – 45 = 4 3.(3x + 4) + (4x – 7) = 11 4.(4x – 10) + (6x +30) = Find the slope of the line k. 6. Find the slope of a perpendicular line to k Standardized Test Practice: y x k A (0,1) (-6,-2) B (6,4) C ACBD 1/22-1/2 -2

5-Minute Check on Algebra Transparency 1-1 Click the mouse button or press the Space Bar to display the answers. 1.6x + 45 = 18 – 3x 2.x 2 – 45 = 4 3.(3x + 4) + (4x – 7) = 11 4.(4x – 10) + (6x +30) = Find the slope of the line k. 6. Find the slope of a perpendicular line to k Standardized Test Practice: 9x +45 = 18 9x = -27 x = -3 x² = 49 x = √49 x = +/- 7 7x - 3 = 11 7x = 14 x = 2 10x + 20 = x = 160 x = 16 ∆y y 2 – y 1 4 – m = = = = = ---- ∆x x 2 – x 1 6 – ACBD 1/22-1/2 -2 y x k A (0,1) (-6,-2) B (6,4) C ∆x ∆y

Objectives Understand the tangent problem Understand the velocity problem

Vocabulary Tangent – a line that touches the curve in only one point; same slope as curve at intersection point Secant – a line that intersects the curve in only two points; average slope (rate of change) between two points Average Velocity – distance traveled divided by time elapsed average of two instantaneous velocities; the slope of the secant line Instantaneous Velocity – the velocity at an instant in time; limit of average velocity as the two points of the secant get closer together (as ∆x →0) the slope of the tangent line

Slope rise ∆y d - b m = = = = rate of change of y with respect to x run ∆x c - a Tangent Problem: Slope of a curve at any point (slope of a tangent at that point) can be estimated by taking the slope of the secant line with those two x-values (a and c) being very close together y = f(x) Secant Line P Q (c,d) (a,b) y x

Summary & Homework Summary: –Two points determine a line –Three noncollinear points determine a plane Homework: pg 9,10: 7-8, 13, 15, 17, 22-23, 32, 34-35